Transcript RC Circuits

RC Circuits
AP Physics C
Montwood High School
R. Casao
RC Circuits
To date, we have studied steady-state direct
current circuits in which the current is
constant.
 In circuits containing a capacitor, the current
varies over time.
 When a potential difference is applied across
a capacitor, the rate at which the capacitor
charges depends on the capacitance and on
the resistance in the circuit.
 An RC circuit can store charge and release it
at a later time.

Charging a Capacitor
Consider a series circuit
containing a resistor and a
capacitor that is initially
uncharged.
 With switch S open, there is
no current in the circuit.
 When switch S is closed at
t = 0 s, charges begin to flow
and a current is present in
the circuit and the capacitor
begins to charge.

Charging a Capacitor
The gap between the capacitor plates represents
an open circuit and charge does not pass from
the positive plate to the negative plate.
 Charge is transferred from one plate to the other
plate through the resistor, switch, and battery
until the capacitor is fully charged.
 The value of the maximum charge depends on
the EMF of the battery.
 Once the maximum charge is reached, the
current in the circuit is zero.

Charging a Capacitor

Applying Kirchhoff’s loop rule to the circuit
after the switch is closed:
E  I  R  Vcapacitor  0
q
E  IR   0
C
I ·R is the potential drop across the resistor.
 q/C is the potential drop across the
capacitor.
 I and q are instantaneous values of the
current and charge as the capacitor
charges.

Charging a Capacitor
At t = 0 s, when the switch is closed, the
charge on the capacitor is 0 C and the initial
current is:
E
I max 
R
 At t = 0 s, the potential drop is entirely
across the resistor.
 As the capacitor is charged to its maximum
value Q, the charges quit flowing and the
current in the circuit is 0 A and the potential
drop is entirely across the capacitor.

Charging a Capacitor
Maximum charge: Q max  C  E
 From t = 0 s until the capacitor is fully
charged and the current stops, the
amount of current in the circuit decreases
over time and the amount of charge on
the capacitor increases over time.
 To determine values for the current in the
circuit and for the charge on the capacitor
as functions of time, we have to use a
differential equation.

Charging a Capacitor – Current Equation

Beginning with Kirchhoff’s loop equation,
differentiate the equation with respect to
time: 
q
d E  I  R  
C

0
dt
dE
0
dt
d I  R 
dI
 R 
dt
dt
dI 1 dq
R  
0
dt C dt
 q
d  
 C    1  dq
dt
C dt
1 dq
dI

 R 
C dt
dt
Charging a Capacitor – Current Equation

Replace dq/dt with I:
1
dI
 I  R 
C
dt

I
dI
 R 
C
dt
Get the current terms on one side of the
equation and the other terms on the other
side of the equation:
dI
-1

 dt
I
R C

Integrate both sides of the equation.
Charging a Capacitor – Current Equation
The limits of integration for the current side
of the equation is from Imax (at t = 0 s) to
the current value at time t.
 The limits of integration for the time side of
the equation is from t = 0 s to time t.

t
dI
-1


dt
Im ax I R  C 0
I
Charging a Capacitor – Current Equation

Left side:
1
I
I
I  dI  ln I I  ln I  ln I max  ln
I
I max
I
max

max
Right side:
t
1
1
1
t
  dt 
t0 
 t  0 
R C 0
R C
R C
t

R C
Charging a Capacitor – Current Equation

Combining both sides of the integration:
ln

Imax
t

R C
To eliminate the natural log term (ln), we
can use the terms as exponents for the
base e. From the properties of logarithms:
e
ln
e
I
I
I max
e
t
R C
ln x
x
simplifies to
I
I max
e
t
R C
Charging a Capacitor – Current Equation

Current in an RC circuit as a function of
time:
t
t
I(t)  I max  e

R C
E RC
 e
R
Graph of Current vs. time for a charging
capacitor:
Charging a Capacitor – Charge Equation

To find the charge on the capacitor as a
function of time, begin with Kirchhoff’s loop
equation:
E  I  R  Vcapacitor  0

Replace I with dq/dt:
q
E  IR   0
C
dq
q
E  R   0
dt
C
Charging a Capacitor – Charge Equation

Get the dq/dt term on one side of the
equation: dq
q
dt

R  E 
C
Divide both sides by R to solve for dq/dt:
dq E
q
 
dt R R  C

Common demominator, R·C:
dq E  C
q


dt R  C R  C
dq E  C  q

dt
R C
Charging a Capacitor – Charge Equation
 Multiply both sides by dt:
 EC  q 
dq  
  dt
 R C 

Divide both sides by E·C-q to get the charge terms
together on the same side of the equation:
dq
dt

EC  q R C

In problems involving capacitance, q is positive, so
multiply both sides by –1 to make the charge
positive:
dq
-1

 dt
 EC  q R C
Charging a Capacitor – Charge Equation

Integrate both sides of the resulting
differential equation.
– For the charge side of the equation, the limits
of integration are from q = 0 at t = 0 s to q at
time t. I rearranged the equation to put the
positive q first followed by the negative E·C.
– For the time side of the equation, the limits of
integration are from 0 s to t.
dq
-1
t
 0
 dt
0
q  EC
R C
q
Charging a Capacitor – Charge Equation

Left side:
dq
q
 ln q  E  C 0  ln q  E  C  ln 0  E  C
0
q  EC
q
q  EC
 ln q  E  C  ln  E  C  ln
 EC

Right side:
1
1 t
1 t
 dt 
 0 dt 
t 0
0
R C
R C
R C
1
t

 t  0  
R C
R C
t
Charging a Capacitor – Charge Equation

Combining both sides of the integrals:
q  EC
t
ln

 EC
R C

To eliminate the natural log term (ln), we
can use the terms as exponents for the base
e.
e
q EC
ln
EC
e
t
RC
Charging a Capacitor – Charge Equation

Simplify:

Solve for q:
q  EC
e
 EC
t
R C
– Multiply both sides by –E·C.
– Add E·C to both sides.
q  EC  ECe

Factor out the E·C:
t
R C
t
R C
q  E  C  (1  e )
Charging a Capacitor – Charge Equation

Substitute: Qmax = E·C
t
R C
q(t)  Q max  (1  e )

Graph of Charge vs. time for a charging
capacitor:
Charging A Capacitor

Current has its
maximum value
of I max = E/R at
t = 0 s and
decays
exponentially to
0 A as t infinity.
Charging A Capacitor

The charge on a
capacitor is 0 C
at t = 0 s and
approaches a
maximum value
of Qmax = C·E as
t infinity.
Charging Capacitor Graphs
current
voltage
charge
Time Constant R·C
The quantity R·C, which appears in the exponential
component of the charge and current equations is
called the time constant t of the circuit.
 The time constant is a measure of how quickly the
capacitor becomes charged.
 The time constant represents the time it takes the:
 current to decrease to 1/e of its initial value.
 charge to increase from 0 C to C·E·(1-e-1) =
0.63·C·E.
 The unit for the time constant is seconds.
Ω · F = (V/A)(C/V) = C/(C/s) = s

Charge and Current during the
Charging of a Capacitor
Qmax
q
0.63 I
Capacitor
Rise in
Charge
t
Time, t
I
i
Capacitor
Current
Decay
0.37 I
t
Time, t
In a time t of one time constant, the charge q
rises to 63% of its maximum, while the current
i decays to 37% of its maximum value.
Discharging a Capacitor

Removing the battery from the circuit while
keeping the switch open leaves us with a
circuit containing only a charged capacitor
and a resistor.
Discharging a Capacitor
When the switch is open, there is a
potential difference of Q/C across the
capacitor and 0 V across the resistor since
I = 0 A.
 If the switch is closed at time t = 0 s, the
capacitor begins to discharge through the
resistor and a current flows through the
circuit.
 At some time during the discharge, current
in the circuit is I and the charge on the
capacitor is q.

Discharging a Capacitor – Charge Equation

To find the charge on the capacitor as a function
of time, begin with Kirchhoff’s loop equation.
There is no E term in the equation because the
battery has been removed. The I·R term is
negative because the energy carried by the
charges is dissipated in the resistor.
 I  R  Vcapacitor  0
q
IR 
C
q
 IR   0
C
Discharging a Capacitor – Charge Equation

Replace I with -dq/dt because the current in
the circuit is decreasing as the capacitor
discharges over time:
dq q
R

dt C

Get the charge terms on one side of the
equation and the remaining variables on the
other side of the equation.
dq
1

 dt
q R C
Discharging a Capacitor – Charge Equation

Integrate both sides of the resulting differential
equation.
– For the charge side of the equation, the limits of
integration are from q = Qmax at t = 0 s to q at time t.
– For the time side of the equation, the limits of
integration are from 0 s to t.

q

q
Q max
Q max
dq t  1
 0
 dt
q
R C
1
1 t
 dq 
 0 dt
q
R C
Discharging a Capacitor – Charge Equation


q
Q max

Left side:
1
 dq  ln q
q
q
Q max
 ln q  ln Q max
q
 ln
Q max
Right side:
1 t
1 t
1
t
 0 dt 
t 0 
 t  0  
R C
R C
R C
R C
Discharging a Capacitor – Charge Equation

Combining both sides of the integrals:
q
t
ln

Q max R  C

To eliminate the natural log term (ln), we
can use the terms as exponents for the
base e.
ln
e
q
Q max
e
t
R C
Discharging a Capacitor – Charge Equation

Simplify:

Solve for q:
q
e
Q max
t
R C
qt   Q max  e
t
R C
Discharging a Capacitor – Current Equation

To find the current on the capacitor as a function
of time, begin with the charge equation.
qt   Q max  e

t
R C
Current I = dq/dt, so take the derivative of the
charge equation with respect to time:

d Q max  e
dq


dt
dt
t
R C



Discharging a Capacitor – Current Equation

Left side:

Right side:
dq
I
dt
d Q max  e

dt
 Q max

e 
d



dq
  Q dt I 

max
dt
 t 
d

-t
-t
 1 dt
R C

R C
R C
e 
 Q max  e 

dt
R  C dt
-t
R C
- Q max R-tC

e
R C
-t
R C
Discharging a Capacitor – Current Equation

Combining the two sides of the integral:
 Q max
I
e
R C

Qmax = C·V, substituting:
 CV
I t  
e
R C
It   I max  e
t
R C
t
R C
t
R C
V

e
R
t
R C
Discharging a Capacitor – Current Equation
The negative sign indicates that the
direction of the discharging current is
opposite to the direction of the charging
current.
 The voltage V across the capacitor is equal
to the EMF of the battery since the
capacitor is fully charged at the time of the
switch is closed to discharge the capacitor
through the resistor.

Discharging Capacitor Graphs
voltage
charge
current
Bonus Equations!
I have never seen these equations in any
textbook and had never been asked to find
the voltage across the capacitor as a
function of time. I got these equations
from the E & M course I took Fall 06.
 Discharging Capacitor:

qt  Q max  e
V t  

C
C
V(t)  E  e
t
R C
t
R C
E  C RtC

e
C
Bonus Equations!

Charging Capacitor:
q t 
V t  

C
Q max

V(t)  E  1  e

t
R C

 1  e

C



t
R C

 EC
t


R C 
 1  e 
C 

Energy Conservation in Charging a
Capacitor
During the charging process, a total
charge Q = EC flows thru the battery.
 The battery does work W = QmaxE or
W = CE2.
 Half of this work is accounted for by the
energy stored in the capacitor:
U = 0.5QV = 0.5QmaxE = 0.5CE2
 The other half of the work done by the
battery goes into Joule heat in the
resistance of the circuit.

– The rate at which energy is put into the resistance
R is:
dWR
2
 I R
dt
– Using the equation for current in a charging
capacitor:
t 2
2
2t


dWR
E RC
E
   e   R 
 e RC
dt
R
R

– Determine the total Joule heat by integrating from
t = 0 s to t = :
2t
2
WR  

0
E
R C
 e  dt
R

Substitute: let
2t
R C
2t
2
x
, then dx 
 dt
R C
R C
R C
dt 
 dx
2
E
E
x R  C
WR  
 e  dt  
e 
 dx
0 R
0 R
2
2
E R  C  x
WR 

  e  dx
0
R
2
2
2

E C
E C
x
x 
WR 
  e  dx 
 e
o
0
2
2

2

2
E C
E C

0
WR 
  e  e 
 0  1
2
2
2
E C
WR 
2
2


2
The result is independent of the resistance R; when
a capacitor is charged by a battery with a constant
EMF, half the energy provided by the battery is
stored in the capacitor and half goes into thermal
energy.
 The thermal energy includes the energy that goes
into the internal resistance of the battery.

Capacitors and Resistors in Parallel
The capacitor in the figure is initially
uncharged when the switch S is closed.
 Immediately after the switch is closed, the
potential is the same at points c and d.

An uncharged capacitor does not resist the flow of
current and acts like a wire.
 No current flows thru the 8 Ω resistor; the
capacitor acts as a short circuit between points c
and d.
 Apply Kirchhoff’s loop rule to the outer loop
abcdefa: 12 V – 4 Ω·I0 = 0; I0 = 3 A

After the capacitor is fully charged, no more charge
flows onto or off of the plates; the capacitor acts
like a broken wire or open in the circuit.
 Apply Kirchhoff’s loop rule to loop abefa:
12 V – 4 Ω·If – 8 Ω·If = 0; 12 V – 12 Ω·If; If = 1 A
